# Inverse Function

• May 22nd 2011, 06:25 AM
Self Inverse

$\displaystyle f(x) = \frac{2}{x } - x$ so i have done $\displaystyle f(f(x)) -x = 0$ so $\displaystyle \frac{2}{\frac{2}{ x} -x} - (\frac{2}{ x} -x) - x = 0$ How does that equal $\displaystyle \frac{4{x}^{ 2} -4}{x(2-{x}^{ 2} ) }$ ???
• May 22nd 2011, 06:30 AM
SpringFan25
Shouldn't it be $\displaystyle f(f^{-1}(x)) - x =0$?

it may be easier to write
$\displaystyle y=\frac{2}{x} - x$
$\displaystyle yx=2 - x^2$

$\displaystyle x=\frac{-y \pm \sqrt{y^2 + 8}}{2}$

and hence the inverse is
$\displaystyle f^{-1}(x)=\frac{-x \pm \sqrt{x^2 + 8}}{2}$
• May 22nd 2011, 06:33 AM
Quote:

Originally Posted by SpringFan25
Shouldn't it be $\displaystyle f(f^{-1}(x)) - x =0$?

who knows what the name is called all i know is i have to do the formula i've put up on there.
• May 22nd 2011, 06:49 AM
SpringFan25
(not sure if you saw my previous post before i edited it to include the answer)

using your formula you should get the same answer as i posted:

$\displaystyle f(f^{-1}(x)) -x =0$
$\displaystyle \frac{2}{f^{-1}(x)} - f^{-1}(x) -x =0$

multiple everything by f^-1(x):
$\displaystyle 2 - \left[f^{-1}(x)\right]^2 -xf^{-1}(x) =0$

this is a quadratic in $\displaystyle f^{-1}(x)$. the coefficients (a,b,c) are (-1,-x,2)

$\displaystyle f^{-1}(x) = \frac{x \pm \sqrt{x^2 + 8}}{-2} = \frac{-x \pm \sqrt{x^2 + 8}}{2}$